Question

The difference of a number and its reciprocal is 1/2.the sum of their squares is

Answer 1

x-1/x=1/2
(x-1/x)^2=(x+1/x)^2-4
(1/2)^2=x^2+1/x^2-2
x^2+1/x^2=9/4
ans=9/4

Answer 2

Given,

Difference between a number and its reciprocal = $\frac{1}{2}$

To find,

Sum of their squares.

Solution,

First of all, let the number be x.

Now, the reciprocal of this number will be $\frac{1}{x}$.

It is given that the difference between the two is $\frac{1}{2}$. So, this can be expressed as,

$x-\frac{1}{x} =\frac{1}{2}$     ...(1)

We have to find the sum of their squares that is,

$x^{2} +(\frac{1}{x} )^{2}$.

This can be determined simply by using the algebraic identity

$(a-b)^2=a^2-2ab+b^2$

On squaring both the sides in equation (1), we get,

$(x-\frac{1}{x} )^2=(\frac{1}{2})^2$

⇒ $x^2-2(x)(\frac{1}{x} )+(\frac{1}{x} )^2=\frac{1}{4}$

Simplifying the above equation, we get,

⇒ $x^2+(\frac{1}{x} )^2 -2=\frac{1}{4}$

⇒ $x^2+(\frac{1}{x} )^2=\frac{1}{4}+2$

⇒ $x^2+(\frac{1}{x} )^2 =\frac{9}{4}$, which is the required value.

Therefore, the sum of "the squares of the number and its reciprocal" will be $\frac{9}{4}$.